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Applied Statistics-Homework 4 Solved
Exercise 1: Mammalian Sleep
We’ll use the msleep dataset from the ggplot2 package for this exercise. At first glance of the msleep data, you may notice some missing values encoded as NAs. For this question, we will use the sleep (sleep_total) and bodyweight of an animal, which have no missing values.
part a
I wonder about how the amount of sleep required by an animal changes based on the bodyweight of an animal. For example, do animals who weigh more require more sleep. What would be the primary purpose of fitting a model like this? Bold your answer below.
(a) predicting an observation (b) explaining a structure/system
part b
Fit a linear model to estimate the sleep required based on the bodyweight. Interpret the slope for this model. # Use this code chunk for your answer.
lm1 = lm(sleep_total ~ bodywt, data = msleep) summary(lm1)
##
## Call:
## lm(formula = sleep_total ~ bodywt, data = msleep)
##
## Residuals:
## Min 1Q Median 3Q Max ## -7.7008 -2.3787 -0.4268 3.2732 9.1731
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.7269205 0.4773797 22.470 < 2e-16 ***
## bodywt -0.0017647 0.0005971 -2.956 0.00409 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.254 on 81 degrees of freedom
## Multiple R-squared: 0.09735, Adjusted R-squared: 0.08621
## F-statistic: 8.736 on 1 and 81 DF, p-value: 0.004085
part c
Calculate the 90% confidence interval for the slope of this model. Note: the multiplier (t∗) is 1.6639 for this
situation. Make sure to show your setup for the calculation. Report the 90% confidence interval below.
# Use this code chunk, if needed, to calculate your answer.
lb = -0.0017647 - (1.6639 * 0.0005971) up = -0.0017647 + (1.6639 * 0.0005971) print(c(lb, up))
## [1] -0.0027582147 -0.0007711853 Answer:
part d
Report and interpret the coefficient of determination for this relationship.
# Use this code chunk (if needed) to perform any necessary analyses.
r_2 = summary(lm1)$r.squared r_2
## [1] 0.09735061
part e
Based on part d, calculate the correlation.
# Use this code chunk (if needed) to perform your analyses.
-sqrt(r_2)
## [1] -0.3120106 part f
For the three-toed sloth, calculate the predicted total amount of sleep and the corresponding residual.
# Use this code chunk (if needed) to perform any necessary analyses or calculations.
msleep = msleep
sloth = subset(msleep, msleep$name == 'Three-toed sloth') sloth[,11]
## # A tibble: 1 x 1
## bodywt ## <dbl>
## 1 3.85
predicted = 10.7269205 + ( -0.0017647 * 3.85) predicted
## [1] 10.72013
observed = sloth[,6] observed
## # A tibble: 1 x 1
## sleep_total
## <dbl>
## 1 14.4
observed - predicted
## sleep_total
## 1 3.679874
Exercise 2: Hand Calculations
We’ve used R to generate the summary statistics for the msleep dataset so far. Now let’s take a moment and confirm some of these calculations “by hand” (still using R to perform the calculations).
part a
Calculate x¯, y¯, Sxx, and Sxy for the msleep dataset. Clearly label and print your results.
# Use this code chunk for your answer.
x_bar = mean(msleep$bodywt) x_bar
## [1] 166.1363
y_bar = mean(msleep$sleep_total) y_bar
## [1] 10.43373
s_xy = sum((msleep$bodywt - x_bar) * (msleep$sleep_total - y_bar)) s_xy
## [1] -89590.99
s_xx = sum((msleep$bodywt - x_bar)ˆ2) s_xx
## [1] 50767575
part b
Calculate the estimates for β0 and β1, using the values calculated in part a. Clearly label and print your results. Compare these estimates to what you found in Exercise 1b.
# Use this code chunk for your answer.
beta_1 = s_xy/s_xx beta_1
## [1] -0.001764729
coef(lm1)[2]
## bodywt
## -0.001764729
beta_0 = y_bar - (beta_1 * x_bar) beta_0
## [1] 10.72692
coef(lm1)[1]
## (Intercept)
## 10.72692
.
part c
Calculate the residuals for the dataset. You can use any built-in R function to generate the fitted values of the dataset, but you should not use a built-in function to calculate the residuals. No need to print all of the residuals, but please do print the first few residuals.
# Use this code chunk for your answer. fit1 = fitted(lm1)
residuals = msleep$sleep_total - fit1 residuals
## 1 2 3 4 5 6 7 ## 1.4613159 6.2739266 3.6754619 4.1731130 -5.6680834 3.6798737 -1.9907612
## 8 9 10 11 12 13 14 ## -3.7268411 -0.6022143 -7.7008025 -5.3678021 -1.3256358 -0.7185380 1.7738207 ## 15 16 17 18 19 20 21 ## -0.4268146 -2.4251558 -1.6269117 6.6792560 -5.4217146 7.2760795 -2.3321569
## 22 23 24 25 26 27 28 ## 8.9731201 -6.9074969 -7.2969163 -0.6255617 0.1907268 4.1732048 1.7789031
## 29 30 31 32 33 34 35 ## -0.9265676 -7.2386736 -6.6151377 -4.3769186 -4.4222881 -2.6175073 -1.2239734
## 36 37 38 39 40 41 42 ## 4.3155834 8.6737324 -0.6149203 3.4731730 3.5732913 2.0731413 1.7731183
## 43 44 45 46 47 48 49 ## 9.1730971 3.8735489 0.2755501 -3.0265499 3.7731289 -2.3225087 -6.8289781
## 50 51 52 53 54 55 56 ## -0.9348017 5.3599608 -0.1504476 3.0580814 -1.2823876 -0.4259499 0.2750207
## 57 58 59 60 61 62 63 ## 0.7731166 2.9759384 -7.0751538 -5.0330722 0.3750207 7.4789632 -5.3205675
## 64 65 66 67 68 69 70 ## 2.2736442 -2.0268429 -1.1256093 -2.3267881 0.5733407 -0.1267052 5.8747030
## 71 72 73 74 75 76 77 ## 3.0732577 5.1734413 2.0731642 -1.4747127 -2.1189792 5.0732771 -5.9607376
## 78 79 80 81 82 83
## 4.8746678 -1.8267370 -5.2210401 -4.4233910 1.7790443 -0.9194557 part d
Calculate the SSR, SSE, and SST for the model. You may use anything that you have calculated in the earlier parts of this question. Clearly label and print these three values.
# Use this code chunk for your answer. sst = sum((msleep$sleep_total - y_bar)ˆ2) sst
## [1] 1624.066
sse = sum((fit1 - msleep$sleep_total)ˆ2) sse
## [1] 1465.962
ssr = sum((fit1 - y_bar)ˆ2) ssr
## [1] 158.1038 part e
How are SSR, SSE, and SST related to each other?
part f
Calculate the coefficient of determination from the values calculated in part d. How does this compare to what you found in Exercise 1d (based on Exercise 1b)?
# Use this code chunk for your answer.
ssr/sst
## [1] 0.09735061
summary(lm1)$r.squared
## [1] 0.09735061
Exercise 3: Professor Beauty
In the article, “Beauty in the classroom: Instructors’ pulchritude and putative pedagogical productivity” in the journal Economics of Education Review (http://www.sciencedirect.com/science/article/pii/S0272775 704001165), Hamermesh and Parker explored how course evaluations may be associated with a professor’s physical appearance. The data were gathered from end of semester student evaluations for a large sample of professors from the University of Texas at Austin. Later, six undergraduate students were asked to look at each professor’s photo and rate their physical beauty from 1 to 10.
We will focus on two variables:
• score is the average student rating that each professor received.
• bty_avg is the average score provided by the six student raters.
part a
Load the data Prof_Evals.csv into your markdown file as prof_evals. To do this, you’ll want to download the csv file into the same Folder as your Homework4.Rmd file, and then use a line of code to call in the csv file. Print the number of professors included in the dataset and the number of variables recorded for each professor.
# Use this code chunk to answer the question.
setwd("~/Desktop/data") prof_evals = read.csv("Prof_Evals.csv") dim(prof_evals)[1] # number of profs
## [1] 463
dim(prof_evals)[2] # number of variables recorded for each professor
## [1] 19 part b
Create a simple linear model with these two variables, named prof_model. Run the summary of your model.
# Use this code chunk to answer the question.
prof_model = lm(score ~ bty_avg, data = prof_evals) summary(prof_model)
##
## Call:
## lm(formula = score ~ bty_avg, data = prof_evals)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9246 -0.3690 0.1420 0.3977 0.9309 ##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.88033 0.07614 50.96 < 2e-16 *** ## bty_avg 0.06664 0.01629 4.09 5.08e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5348 on 461 degrees of freedom
## Multiple R-squared: 0.03502, Adjusted R-squared: 0.03293 ## F-statistic: 16.73 on 1 and 461 DF, p-value: 5.082e-05 part c
In RMarkdown, you can directly input code into your text space (with a certain formatting) and then have the knitted file convert that code into the number produced by that code!
I can extract a particular value from this model into my text space by doing the following:
• Type r, a space, and then your code
• Then surround this all with backtick marks (the character that you use to create a code chunk, typically found on the top left of your keyboard).
An example follows. For this to run, remove the two # signs, one in the R chunk and the other in the line of text immediately following the chunk:
age_model = lm(score ~ age, data = prof_evals)
βˆ0 = 4.4619324
Knit the pdf, and you’ll notice that this is printed as a value! The mathematical symbol before is surrounded by dollar signs, since that is typed using TeX style. The part after the equals sign is the code transformed to its value!
Now... for this exercise, report the values of the intercept and slope for your model in the space below using the automated process.
part d
Interpret βˆ0, βˆ1, and R2 in context with the appropriate value plugged in (you don’t have to use R code for the interpretation statements, but you can!)
Exercise 4: Professor Beauty Inference part a
Calculate the expected evaluation score values for professors with average beauty scores of 6.25 and 9.5.
# Use this code chunk for your answer.
as.numeric(coef(prof_model)[1] + (coef(prof_model)[2] * 6.25))
## [1] 4.296814
as.numeric(coef(prof_model)[1] + (coef(prof_model)[2] * 9.5))
## [1] 4.513384
part b
Calculate an 95% confidence interval for the true slope.
Print both the lower and upper bound. Make sure these are clearly labeled.
# Use this code chunk for your answer.
lb = 0.06664 - (2 * 0.01629) ub = 0.06664 + (2 * 0.01629)
print(c(lb,up))
## [1] 0.0340600000 -0.0007711853 part c
For the default hypothesis test for the intercept, report the following:
• The null and alternative hypotheses for the slope (in symbols)
• The value of the test statistic
• The p-value of the test
• A statistical decision at α = 0.05
No need to extract these specific values in your code. Simply observe the values in the output and report this information in text below.
# Use this code chunk for your answer, if needed.
part d
Suppose that the University of Illinois had previously found that for each increase in the student perceived average beauty rating of a professor of 1 on a scale from 1 to 10, that the estimated average student rating score of that professor would increase by 0.1, on average.
Might this same relationship be true for the University of Texas at Austin, where our data came from? Or is there evidence that this relationship for the University of Texas is different from the University of Illinois? In other words, use a t-test to test:
• H0 : β1 = 0.1 • H1 : β1 = 06 .1
Calculate and report the value of your test statistic.
# Use this code chunk for your answer.
t_stat = (0.06664 - 0.1)/0.01629
t_stat
## [1] -2.047882
part e
If a news article used this data to make the claim “Professors get a bump in their evaluation scores if they’re viewed as attractive by students!” would you agree with that conclusion? Why or why not?
There is not an objectively right answer for this question. Consider all of the information that we’ve compiled and calculations that we’ve performed to help guide your response.
Exercise 5: Formatting
The last five points of the assignment will be earned for properly formatting your final document. Check that you have:
• included your name on the document
• properly assigned pages to exercises on Gradescope
• select page 1 (with your name) and this page for this exercise (Exercise 5)
• all code is printed and readable for each question
• generated a pdf file
